Question

a train starting from rest picks up a speed of 20 metre per second in 200 second it continues to move at the same rate for next five hundred seconds and then brought to rest in another 800 second find the uniform rate of acceleration,uniform rate of retardation,total distance covered before stopping, average speed

Answer 1

Answer:

(a)Uniform rate of acceleration=0.1 m/s2.

(b)Uniform retardation  = 0.2 m/s2.

(c) Total distance covered before stopping =  13 km

(d) Average speed = 16.25 m/s

Explanation:

First let us plot the graph with the data given.

Let the starting point be the origin in the speed - time graph.

(a) Uniform rate of acceleration  = slope of AB

                                               = (20−0) (m/s)/ [(200−0) s]

                                               = 0.1 m/s2.

(b) Uniform retardation = − slope of CD

                                = − (0−20)(m/s) /[(800−700)s]

                                = 0.2 m/s2.  

(c) Total distance covered before stopping = area under the curve ABCD

                                                            = area of ΔABE + area of rectangle BCFE + area of ΔCDF

                                                            = ½ x 200 x 20 + 500 x 20 +  ½ x 100 x 20  m

                                                            = 2000 + 10000 + 1000   m

                                                            = 13,000 m = 13 km

(d) Average speed = Total distance covered/total time taken = 13000 m /(800 s)

                                                                                      = 16.25 m/s